a. Explanation Given data: Wavelengths of absorption spectrum is 200 nm, 300 nm and 500 nm. Formula used: Write the expression for energy. E photon = h c λ (1) Here, E photon is the energy, h is the Planck’s constant, which has the standard value as 6.62 × 10 − 34 m 2 ⋅ kg s , c is the speed of the light 3 × 10 8 m s , and λ is the wavelength. Explanation: For 200 nm, substitute 6.62 × 10 − 34 m 2 ⋅ kg s for h , 3 × 10 8 m s for c and 200 × 10 − 9 m for λ in equation (1), E photon = ( 6.62 × 10 − 34 m 2 ⋅ kg s ) × ( 3 × 10 8 m s ) 200 × 10 − 9 m = ( 1.986 × 10 − 25 200 × 10 − 9 ) J = 9.93 × 10 − 19 J Consider the conversion factor of J to eV. 1 J = ( 1 1.6 × 10 − 19 ) eV Consider the expression of energy in eV. E photon = 9.93 × 10 − 19 J 1.6 × 10 − 19 J = 6.206 eV ≅ 6 .21 eV For 300 nm, substitute 6.62 × 10 − 34 m 2 ⋅ kg s for h , 3 × 10 8 m s for c and 300 × 10 − 9 m for λ in equation (1), E photon = ( 6.62 × 10 − 34 m 2 ⋅ kg s ) × ( 3 × 10 8 m s ) 300 × 10 − 9 m = ( 1 b. To determine Wavelength’s of the emission spectrum.

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Chapter 29, Problem 57GP

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To draw: Atom’s energy level diagram.

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Wavelength’s of the emission spectrum.

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Chapter 29, Problem 56GP

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